The amount of Ag (atomic mass = 108) deposited at the cathode when a current of 0.5 amp is passed through a solution of AgNO $_{3}$ for 1 hour is closest to :

Q. A current of

0.5

ampere when passed through

A g {N O}_{3}

solution for

193

seconds deposited

0.108 g

A g

. The equivalent weight of

A g

is:

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Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO3 for 15 minutes.

The correct option is A 2.014Ag++e−→Agmol of e− passed=change96500=2×15×6096500=0.019 mol1 mol→108 g0.19mol→108×0.01g =2.014g of AgHence, the correct option is A

Calculate the mass of $A g$ deposited at cathode when a current of 2 amperes was passed through a solution of $A g N O_{3}$ for 15 minutes.

The correct option is A $2.014$
${A g}^{+} + e^{-} \to A g$

$m o l o f e^{-} p a s s e d = \frac{c h a n g e}{96500} = \frac{2 \times 15 \times 60}{96500} = 0.019 m o l$

$1 m o l \to 108 g$

$0.19 m o l \to 108 \times 0.01 g$

$= 2.014 g$ of $A g$

Hence, the correct option is $A$